A more succinct group object diagram (all axioms in one connected diagram), questions about its properties... Here is the definition of group object from nLab.  They give 3 associated maps $* \xrightarrow{1} G$, $m: G^2 \to G$, and $-^{-1}: G \to G$ and require 3 commutative diagrams to complete the axioms for a "group".
Here, however, I've managed to condense these down into one (connected) diagram:

We say that $G \in \text{Ob}(C)$ (a category with binary products and a terminal object $*$) forms a group object precisely when the above diagram commutes.
Note that $1\times \text{id}$ is notation implying that we use the map $1 =$ the composition of the two maps $(G \xrightarrow{!} * \xrightarrow{1} G)$, and the appropriate $\Delta$ is used wherever it is needed.  I like the simpler notation that I've used.
Anyway, this diagram is more economical than the combined other 3 since we re-use arrows such as $m\times \text{id}$ at least twice (two commutative shapes described - triangles or squares).
So, that's background and what I've done on the problem so far.  My questions are:

Can we define a group object to be an object $G$ such that the above diagram commutes when we replace each $m \times \text{id}, 1\times \text{id}, \text{id}\times -^{-1} \times \text{id}$ with  general morphisms $f, g, h$ in the category $C$ ($G \in \text{Ob}(C)$)?  Or can you only recover a group object if you define these edges using products, the diagonal morphism, etc, as shown?


In either case, is $G$ the limit of a commutative square (the outer one), since a cone seems to be formed.

 A: Group theory is algebraic, which means that it is a Lawvere Theory. This essentially means that the category of groups is precisely given by the functor category $[D,Set]_{\text{f.p.p.}}$ of finite product preserving functors, where $D$ is the diagram you have written. In particular, the functors must preserve the terminal object and products.
A: The answer to the first question is no. To see this more clearly, note that asserting that a morphism is of the form, say, $m\times\mathrm{id}$, means 1) that the codomain is equipped with two projection maps, which composed with the given morphism yield $m$ and $\mathrm{id}$, and 2) that the two projection maps form a limiting cone (or rather, are supposed to be sent to a limiting cone; a diagram in which certain cones are supposed to be sent to limiting cones is known as a sketch). Thus, your diagram is strictly speaking incomplete: in full detail, it would include the projection morphisms that define the product structures, and the prescriptions of the corresponding compositions with them.
The answer to the second question is also no. This is because the commutative square looks like a cone (with vertex $G^3$) over a diagram ($G^2\xrightarrow mG\xleftarrow mG^2$). Then a cone over the outer square, considered as a given cone over a diagram, is the same data as a cone over that diagram that factors through the given cone. Consequently, the limiting cone over the outer square is simply the cone with vertex the upper-right corner of the square (the vertex of the square considered a sa cone), with the identity map to itself and the rest of the square's morphisms (morphisms in the diagram) to the other objects. More generally, any cofiltered diagram (any pair of morphism $B\to D\leftarrow C$ in the diagram can be completed to a commutative square by morphisms $B\leftarrow A\to C$, and any parallel pair of morphisms $B\to C$, $B\to C$, can be completed to a fork by a morphism $A\to B$) that is finite looks like a cone over a diagram, hence has a limit given by the vertex of that cone.
